Question

analyze this: a rightward force of 56.1 n is applied to a 7.87 - kg object to accelerate it to the right at 4.47 m/s/s. complete the diagram.
tap on a field to enter or edit its value.
units
force: n
mass: kg
acceln: m/s/s
μ: no unit

Explanation:

Step1: Calculate \(F_{grav}\)

Using \(F_{grav}=mg\) (\(g = 9.8\ m/s^{2}\)), \(F_{grav}=7.87\times9.8=77.126\ N\)

Step2: Calculate \(F_{norm}\)

Since there is no vertical acceleration, \(F_{norm}=F_{grav}\), so \(F_{norm}=77.126\ N\)

Step3: Calculate \(F_{net}\)

Using \(F_{net}=ma\), \(F_{net}=7.87\times4.47 = 35.1789\ N\)

Step4: Calculate \(F_{frict}\)

Using \(F_{net}=F_{app}-F_{frict}\), so \(F_{frict}=F_{app}-F_{net}\). Given \(F_{app}=56.1\ N\), \(F_{frict}=56.1 - 35.1789=20.9211\ N\)

Step5: Calculate \(\mu\)

Using \(F_{frict}=\mu F_{norm}\), so \(\mu=\frac{F_{frict}}{F_{norm}}=\frac{20.9211}{77.126}\approx0.271\)

Answer:

\(F_{norm}=77.126\ N\), \(F_{grav}=77.126\ N\), \(m = 7.87\ kg\), \(a = 4.47\ m/s^{2}\), \(F_{net}=35.1789\ N\), \(F_{frict}=20.9211\ N\), \(\mu\approx0.271\), \(F_{app}=56.1\ N\)